What Is the Acceleration in an Atwood Machine with Blocks of 396 N and 180 N?

[wallace13]

In the drawing, there are two blocks connected by a rope and pulley. One block is on the table while the other is hanging off of the table. Block A, with a weight of 396 N, is on the table. Block B is hanging off of the table, and has a weight of 180 N. Ignore all frictional effects and assume the pulley to be massless.

Find the acceleration of the two blocks.


T= ma
N-Fg= 0
T-Fg=-ma


I made an x and y chart for each of the blocks to show which forces pulled in which direction.

Block A- In the x-direction, there is no Fg or N, but there is T, and the ma is positive. So, I found the equation T= ma... In the y-direction Fg is negative, and there is an N, while ma=0. So I found the equation N-Fg=0.

Block B- There are no forces in the x-direction. In the y-direction, Fg is negative, there is T, and ma is negative, as well. So, I found the equation T- Fg= -ma


I found the mass of block B to be 18.349 kg (I divided 180 by 9.81).

And I substituted all of my known values into my equations...

T-180= -18.349a

I know that T= ma, so I substituted ma for T on the left side of the equation...

18.349a=-18.349a+180

I solved for acceleration and got 4.905, which is the incorrect answer.

 

[avenkat0]

First treat the masses as one system... then try to find the Net force acting on the system... then you will be able to find acceleration

 

[thomate1]

I would approach this Atwood Machine problem by first identifying the known values and setting up free body diagrams for each block. From the given information, we know that the weights of the two blocks are 396 N and 180 N, respectively. Additionally, we are told to ignore frictional effects and assume the pulley to be massless.

Using the equation T=ma, we can set up equations for each block in the x and y directions. In the x-direction, there are no forces acting on either block. In the y-direction, for block A, the only force is its weight (Fg) acting downwards, and for block B, the only force is the tension (T) from the rope acting upwards.

Therefore, our equations become:

Block A: N – Fg = ma
Block B: T – Fg = ma

Substituting in the known values for mass and weight, we get:

Block A: N – 396 = ma
Block B: T – 180 = ma

Since the two blocks are connected by the same rope, the tension in the rope will be the same for both blocks. This means we can set the two equations equal to each other:

N – 396 = T – 180

Solving for N, we get N = T + 216

Substituting this value for N into our equation for block A, we get:

T + 216 – 396 = ma

Rearranging the equation, we get:

T = ma + 180

Now, we can substitute this value for T into our equation for block B:

ma + 180 – 180 = ma

Solving for acceleration, we get:

ma = ma

This means that the acceleration of both blocks is equal and has the value of a = 0. This makes sense, as the system is in equilibrium and there is no net force acting on either block.

In conclusion, the acceleration of the two blocks in this Atwood Machine problem is 0 m/s^2. This means that both blocks will remain stationary and the tension in the rope will remain constant throughout the system.